STRUCTURE OF NITROGEN ISOTOPES
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) (August 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories, which could not lead to the nuclear structure. Under this physics crisis and using the charged UP and DOWN quarks , discovered by Gell-Mann and Zweig, I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” (2003), which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838.68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). In this photo I present the electromagnetic laws governing the nuclear structure, but a student of Einstein (Dr Th. Kalogeropoulos ) criticised my discovery of nuclear force and structure by believing that the nuclear structure is due to the invalid relativity. In fact, here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. You can see my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS . Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications. Natural nitrogen (N) consists of two stable isotopes, nitrogen-14, which makes up the vast majority of naturally occurring nitrogen, and nitrogen-15. Fourteen radioactive isotopes (radioisotopes) have also been found so far, with atomic masses ranging from 10 to 25, and one nuclear isomer, 11mN. All of these radioisotopes are short-lived, with the longest-lived one being nitrogen-13 with a half-life of 9.965 minutes. All of the others have half-lives below 7.15 seconds, with most of these being below five-eighths of a second. Most of the isotopes with atomic mass numbers below 14 decay to isotopes of carbon, while most of the isotopes with masses above 15 decay to isotopes of oxygen. The shortest-lived known isotope is nitrogen-10, with a half-life of about 2.3 microseconds. In order to reveal the structure of nitrogen isotopes I start with the diagram of my paper STRUCTURE OF C-12, C-14 AND N-14 . In the following diagram of the stable N-14 you see that the 6 nucleons of the first horizontal plane have positive spins (+HP1), the 6 nucleons of the second horizontal plane have negative spins (-HP2), and the 2 nucleons of the third horizontal plane have positive spins (+HP3). ' DIAGRAM OF N-14' ' p7(+1/2) ' ' n7(+1/2) +HP3' ' p2(-1/2)..........n4 (-1/2)……….p6(-1/2) ' ' n2 (-1/2)..........p4 (- 1/2)……….n6(-1/2) -HP2 ' ' n1(+1/2)..........p3(+1/2)……….n5(+1/2) ' ' p1(+1/2)..........n3(+1/2)……….p5(+1/2) +HP1 ' ' STRUCTURE OF N-11, N-12 AND N-13' In the absence of three neutrons we get the structure of N-11. In this case the absent n1(+1/2), n2(-1/2) and n6(-1/2) give S =-1/2. Thus the total spin S = +1/2 is given by S = +1 -1/2 = +1/2 However in the absence of two neutrons of opposite spins we get the N-12 with S = +1. Then in the absence of one neutron of positive spin we expect to get the nuclide N-13 of S =+1/2 . Since it has a total spin S = -1/2 we conclude that at the same time the neutron n1(+1/2) changes its spin from S = +1/2 to S = -1/2 in order to make a single horizontal bond with the p4(-1/2). ' ' STRUCTURE OF N-15 WITH S = -1/2 Of course the N-14 has S =+1 because it has the n7(+1/2) and the p7(+1/2) of the third plane, while the nucleons of the first and the second plane give S =0. However In the presence of an extra neutron we get the N-15 of S =-1/2 . In fact, it has the structure of oxygen when the p1(+1/2) is absent. (See my STRUCTURE OF O-16 AND O-15 ).. Especially it consists of 8 neutrons with opposite spins and of 7 protons giving a total S = -1/2. STRUCTURE OF N-16 WITH S = -2 As in the case of C-14 here we have the two extra neutrons of opposite spins like n8(-1/2) and the n9(+1/2) which makes a horizontal bond with the p3(+1/2). However because of the total spin S =-2 we do not follow the structure of N-14. In this case the nucleons n7(+1/2) and p7(+1/2) of the third plane (+HP3) change the spin from S = +1 to S =-1 because they are moved to make a new deuteron below the first horizontal plane in a new third plane with negative spins ( -HP3). Since the total spin of N-16 is S =-2 we also suggest that the n5(+1/2) changes the spin from S =+1/2 to S=-1/2 because it is moved from the +HP1 to -HP2 in order to make a single horizontal bond with p6(-1/2). Since this change gives S=-1 we get S = -1 -1 = -2. As in the case of C-14 in N-16 the nn repulsions of the two extra neutrons lead to the beta decay. ' ' STRUCTURE OF N-17 WITH S =-1/2 As in the case of N-16 the deuteron n7p7 is under the first horizontal plane with S =-1. Since the two extra neutrons like n8(-1/2) and n9(+1/2) give S=0 we conclude that the third extra neutron like the n9(+1/2) makes a single horizontal bond with p1(+1/2). Under this condition we get S= -1+1/2 = -1/2. STRUCTURE OF N-18 WITH S = -1 In the presence of 4 extra neutrons of opposite spins we get the N-18 of S = -1 Since here the deuteron p7n7 is under the first horizontal plane with S =-1. STRUCTURE OF N-19 WITH S = -1/2 Similarly in the presence of 5 extra neutrons with a total spin S =+1/2 we get the structure of N-19 of S =-1/2 because the deuteron p7n7 is under the first horizontal plane with S = -1. Category:Fundamental physics concepts